Q33 of 168 Page 7

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 kmph more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Let the speed of the train and the time taken to cover the same be ‘X’ km/hr and ‘Y’ hrs respectively.



As per the question,


XY = 90 –––––– (i)


(X + 15)(Y – 0.5) = 90 ––––– (ii)


LHS = RHS


Equating the LHS of both equations, and simplyifying it further


XY = XY – 0.5X + 15Y – 7.5


0.5X – 15Y + 7.5 = 0 ––––––– (iii)


Multiplying the above equation by 10,


5X – 150Y + 75 = 0


Simplyfying it further,


X – 30Y + 15 = 0


Putting the value of Y, in equation (iii)



X2 + 15X – 2700 = 0


On applying Sreedhracharya formula







X = – 60 or X = 45


the original speed of train is 45 km/hr as speed can’t be negative.


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